347. Top K Frequent Elements

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本文介绍了一种高效算法，用于从非空整数数组中找出出现频率最高的k个元素。该算法采用unordered_map记录每个元素的出现次数，并利用priority_queue进行筛选，确保时间复杂度优于O(nlogn)。

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Given a non-empty array of integers, return the k most frequent elements.

For example,

Given [1,1,1,2,2,3] and k = 2, return [1,2].

Note:

You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
Your algorithm’s time complexity must be better than O(n log n), where n is the array’s size.

使用unordered_map和priority_queue，时间复杂度O(n*lg(n-k))。

class Solution {
public:
    vector<int> topKFrequent(vector<int>& nums, int k) {
        unordered_map<int, int> ump;
        for(auto num : nums)
            ump[num]++;

        vector<int> res;
        priority_queue<pair<int, int>> pq;
        for(auto it : ump){
            pq.push(make_pair(it.second, it.first));  //pair<first, second>, in priority_queue ths first is frequency, second is number
            if(pq.size() > ump.size() - k){
                res.push_back(pq.top().second);
                pq.pop();
            }
        }
        return res;
    }
};